Indicate the concentration of each ion present in the solution formed by mixing the following.?

Question

(Assume that the volumes are additive.)

(c) 3.50 g of NaCl in 58.6 mL of 0.425 M CaCl2 solution
Na+___M

Ca2+__M

Cl -____M

Answer 1

Since the question states “assume the volumes are additive”, this must be taken into account.

The density of NaCl is 2.165 g/mL so the volume of 3.50g of NaCl = 1.62 mL. Most likely this volume will change slightly when dissolved in water, but no information is given in the problem to account for this possible effect.

The total volume of the resulting solution will be 1.62mL + 58.6 mL = 60.2mL = 0.0602L

The total moles of sodium ions = 3.50g NaCl / 58.44g/mol NaCl = 0.0599 mol Na+
The total moles of calcium ions = 0.0586L x 0.425M = 0.0249 mol Ca2+
The total moles of chloride ions is 0.0599 + 2(0.0249) = 0.110 mol Cl-

The resulting concentrations are:
0.0599 mol Na+ / 0.0602L = 0.995 M Na+
0.0249 mol Ca2+ / 0.0602L = 0.414 M Ca2+
0.110 mol Cl- / 0.0602L = 1.83 M Cl-